Calculate the pH of a 0.81 M Solution of C2H5NH3Cl
Use this premium acid-base calculator to find the pH of ethylammonium chloride solutions, review the weak-acid equilibrium steps, and visualize the resulting hydrogen ion and hydroxide ion concentrations with a responsive chart.
Ethylammonium Chloride pH Calculator
Expert Guide: How to Calculate the pH of a 0.81 M Solution of C2H5NH3Cl
To calculate the pH of a 0.81 M solution of C2H5NH3Cl, you need to recognize what kind of compound this is and how it behaves in water. C2H5NH3Cl is ethylammonium chloride, the salt formed when the weak base ethylamine, C2H5NH2, reacts with hydrochloric acid. Because hydrochloric acid is a strong acid and ethylamine is a weak base, the cation C2H5NH3+ remains acidic in water. That means the solution is not neutral, even though it is a salt solution. Instead, it undergoes hydrolysis and produces hydronium ions.
This is one of the most common acid-base equilibrium problems in general chemistry. Students often expect salts to be neutral, but salts derived from a weak base and a strong acid produce acidic solutions. The correct strategy is to treat ethylammonium ion as a weak acid, determine its acid dissociation constant, and then solve for hydronium concentration. Once you know [H+] or [H3O+], calculating pH is straightforward.
Step 1: Identify the acidic species in solution
When C2H5NH3Cl dissolves in water, it dissociates almost completely:
C2H5NH3Cl → C2H5NH3+ + Cl–
The chloride ion is the conjugate base of a strong acid, so it does not significantly react with water. The important species is the ethylammonium ion, C2H5NH3+, which is the conjugate acid of ethylamine. That ion donates a proton to water according to the equilibrium:
C2H5NH3+ + H2O ⇌ C2H5NH2 + H3O+
This reaction is what makes the solution acidic.
Step 2: Convert Kb of ethylamine into Ka of ethylammonium ion
Most chemistry references list the base dissociation constant for ethylamine rather than the acid dissociation constant for ethylammonium ion. At 25°C, a common textbook value for the base constant of ethylamine is about 5.6 × 10-4. Since conjugate acid-base pairs are related by the water ion-product expression, use:
Ka × Kb = Kw
At 25°C, Kw = 1.0 × 10-14, so:
Ka = (1.0 × 10-14) / (5.6 × 10-4) = 1.79 × 10-11
This tells us ethylammonium ion is a weak acid, which means it only partially ionizes in water. That is why the pH is acidic but not extremely low.
| Chemical quantity | Typical value at 25°C | Why it matters |
|---|---|---|
| Kb for C2H5NH2 | 5.6 × 10-4 | Measures how strongly ethylamine acts as a weak base |
| Kw for water | 1.0 × 10-14 | Connects conjugate acid and base strengths in water |
| Ka for C2H5NH3+ | 1.79 × 10-11 | Determines hydronium ion production from the salt |
| pKa for C2H5NH3+ | About 10.75 | Shows the conjugate acid is weak, but still acidic |
Step 3: Set up the equilibrium expression
The initial concentration of C2H5NH3+ is 0.81 M, assuming complete dissolution of the salt. Let x represent the concentration of hydronium ions produced at equilibrium. Then the ICE setup is:
- Initial: [C2H5NH3+] = 0.81, [C2H5NH2] = 0, [H3O+] = 0
- Change: -x, +x, +x
- Equilibrium: [C2H5NH3+] = 0.81 – x, [C2H5NH2] = x, [H3O+] = x
The acid dissociation expression becomes:
Ka = [C2H5NH2][H3O+] / [C2H5NH3+] = x2 / (0.81 – x)
Substitute the Ka value:
1.79 × 10-11 = x2 / (0.81 – x)
Step 4: Solve for x
Because Ka is very small relative to the concentration, the weak-acid approximation is excellent here. If x is much smaller than 0.81, then 0.81 – x is essentially 0.81. That gives:
x2 = (1.79 × 10-11)(0.81)
x2 = 1.45 × 10-11
x = 3.81 × 10-6
So the hydronium ion concentration is approximately 3.81 × 10-6 M.
If you solve the full quadratic equation instead of using the approximation, you get nearly the same answer. In fact, the difference is negligible for practical chemistry calculations because the percent ionization is extremely small.
Step 5: Convert hydronium concentration to pH
Now apply the pH formula:
pH = -log10(3.81 × 10-6)
pH ≈ 5.42
That is the final answer for the pH of a 0.81 M solution of C2H5NH3Cl when using Kb = 5.6 × 10-4 at 25°C.
Why the pH is not close to 7
Many learners wonder why this salt solution is acidic when chloride itself is neutral. The reason is that acid-base behavior depends on both ions. Chloride is indeed neutral, but ethylammonium is the conjugate acid of a weak base, and it hydrolyzes in water. The acidic contribution from C2H5NH3+ dominates the solution chemistry. Since the parent base ethylamine is reasonably basic, its conjugate acid is only weakly acidic, which is why the pH is moderately acidic rather than strongly acidic.
Common mistakes when solving this problem
- Treating the salt as neutral. Not all salts give pH 7. Salts from weak bases and strong acids give acidic solutions.
- Using Kb directly instead of Ka. The reacting species in water is C2H5NH3+, so you need Ka for the acid form.
- Forgetting the Kw relationship. Ka = Kw / Kb is the key conversion step.
- Assuming a strong acid calculation. A 0.81 M strong acid would have a pH near 0.09, which is completely different from this weak acid salt.
- Rounding too early. Keep enough significant figures until the last step.
How concentration changes the pH
The pH of ethylammonium chloride depends on the concentration. Higher concentrations generally produce more hydronium ions and lower the pH, but because this is a weak acid system, the pH does not decrease in a linear way. The relationship is controlled by the square root dependence from the weak acid approximation.
| Concentration of C2H5NH3Cl | Approximate [H3O+] | Approximate pH | Interpretation |
|---|---|---|---|
| 0.010 M | 4.23 × 10-7 M | 6.37 | Slightly acidic |
| 0.10 M | 1.34 × 10-6 M | 5.87 | Clearly acidic |
| 0.81 M | 3.81 × 10-6 M | 5.42 | Moderately acidic weak-acid salt solution |
| 1.00 M | 4.23 × 10-6 M | 5.37 | Only slightly more acidic than 0.81 M |
Comparison with other salt solutions
It helps to compare C2H5NH3Cl with other common salts:
- NaCl: neutral, because it comes from a strong acid and strong base.
- NH4Cl: acidic, because ammonium is the conjugate acid of weak base ammonia.
- CH3COONa: basic, because acetate is the conjugate base of weak acid acetic acid.
- C2H5NH3Cl: acidic, because ethylammonium is the conjugate acid of weak base ethylamine.
This classification method is one of the fastest ways to predict whether a salt solution will be acidic, basic, or neutral before doing any formal calculation.
When should you use the quadratic method?
The quadratic method is the most rigorous approach because it solves the full equilibrium expression without approximation. In this problem, the approximation is valid because x is tiny compared with 0.81 M. The percent ionization is far below 5 percent, so the simplified method works very well. However, if the solution were extremely dilute or if the acid were much stronger, the quadratic solution would be safer. That is why this calculator lets you choose between the exact quadratic route and the weak acid approximation.
Useful references for acid-base constants and aqueous equilibrium data
For students, instructors, and lab practitioners who want to verify acid-base concepts and water equilibrium assumptions, the following sources are especially useful:
- LibreTexts Chemistry for acid-base equilibrium explanations and worked examples.
- U.S. Environmental Protection Agency (.gov) for water chemistry context and pH significance in environmental systems.
- National Institute of Standards and Technology (.gov) for measurement standards and chemical data practices.
- Massachusetts Institute of Technology Chemistry (.edu) for academic chemistry resources and equilibrium instruction.
Practical interpretation of the result
A pH of about 5.42 means the solution is acidic enough to be clearly below neutral, but nowhere near the acidity of a strong mineral acid at the same formal concentration. This distinction matters in laboratory procedures. If you were preparing reaction media, buffer systems, or salt solutions for kinetics or analytical work, knowing whether a solution is mildly acidic or strongly acidic affects reagent compatibility, indicator selection, corrosion expectations, and protonation state of dissolved amines.
In addition, this problem demonstrates a broader chemistry principle: the pH of a salt solution depends on the acid-base strength of the ions produced, not simply on the fact that a salt is present. Once you can identify the conjugate acid or conjugate base and connect Kb to Ka, these calculations become systematic and much easier to solve.
Short summary
- C2H5NH3Cl dissociates into C2H5NH3+ and Cl–.
- Cl– is neutral, but C2H5NH3+ is a weak acid.
- Use Ka = Kw / Kb = (1.0 × 10-14) / (5.6 × 10-4) = 1.79 × 10-11.
- Solve Ka = x2 / (0.81 – x) for x.
- Find [H3O+] ≈ 3.81 × 10-6 M.
- Calculate pH = -log10[H3O+] ≈ 5.42.
If you are solving the exact question, “calculate the pH of a 0.81 M solution of C2H5NH3Cl,” the correct chemistry-based answer is pH ≈ 5.42 under standard 25°C conditions using the commonly accepted Kb value for ethylamine.